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MySQL ORDER BY ASC and display NULLs at the bottom?
For this, use CASE statement with ORDER BY. Let us first create a table −
mysql> create table DemoTable1937 ( Name varchar(20) ); Query OK, 0 rows affected (0.00 sec)
Insert some records in the table using insert command −
mysql> insert into DemoTable1937 values('Chris');
Query OK, 1 row affected (0.00 sec)
mysql> insert into DemoTable1937 values(NULL);
Query OK, 1 row affected (0.00 sec)
mysql> insert into DemoTable1937 values('Adam');
Query OK, 1 row affected (0.00 sec)
mysql> insert into DemoTable1937 values('John');
Query OK, 1 row affected (0.00 sec)
mysql> insert into DemoTable1937 values('');
Query OK, 1 row affected (0.00 sec)
mysql> insert into DemoTable1937 values(NULL);
Query OK, 1 row affected (0.00 sec)
mysql> insert into DemoTable1937 values('Bob');
Query OK, 1 row affected (0.00 sec)
Display all records from the table using select statement −
mysql> select * from DemoTable1937;
This will produce the following output −
+-------+ | Name | +-------+ | Chris | | NULL | | Adam | | John | | | | NULL | | Bob | +-------+ 7 rows in set (0.00 sec)
Here is the query to ORDER BY ASC and display NULLs at the bottom:
mysql> select * from DemoTable1937 order by case when Name IS NULL then 100 when Name='' then 101 else 103 end desc , Name asc;
This will produce the following output −
+-------+ | Name | +-------+ | Adam | | Bob | | Chris | | John | | | | NULL | | NULL | +-------+ 7 rows in set (0.00 sec)
Updated on: 2019-12-30T07:54:41+05:30
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